Well, I happened across this soln for the case a<b in a book. The author of the book says "By the way, this is a hard problem by any competition standard". Anyway here goes:
We have already seen in my previous post that a<b means an<bn for all n.
Part 1:
Now an+1/bn+1 = an+1/
an+1bn = (an+1/bn ) = (an+bn)/2bn = (1+an/bn)/2
Since an/bn<1 Let an/bn = cos
Then an+1/bn+1 = (1+cos)/2 = cos/2
Hence if a/b = cos
then an/bn = cos(/2n)
Part 2:
Now consider
(bn+1)2-(an+1)2 = (an+bn)/2 (bn-an+bn/2) = (bn2 - an2)/4
Hence we get (bn2 - an2) = (b2-a2)/2n.
which gives bn(1 - an2/ bn2) = bn sin/2n = (b2-a2)/2n
In the limit n
, if limit of the sequence {bn} is is l Then l/2n = (b2-a2)/2n or l = (b2-a2)/ where = cos-1(a/b)
Hence limit {an} = limit {bn} = (b2-a2)/cos-1(a/b)
Phew!
One example, consider a sequence x0 = a<1, xn = (1+xn-1)/2 find the limit of 2n(1-xn2).
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
821
